Determine whether the integral is convergent or divergent. $\int^{\infty}_1 81\frac{\ln(x)}{x}dx$ Determine whether the integral is convergent or divergent.$\int^{\infty}_1 81\frac{\ln(x)}{x}dx$
My answer is $\infty-0=\infty$
But I am unsure whether it is convergent or divergent. 
 A: The constant $81$ is irrelevant, so we can just consider
$$
\int_{1}^{\infty}\frac{\ln x}{x}\,dx
$$
The integral converges if and only if
$$
\int_{e}^{\infty}\frac{\ln x}{x}\,dx
$$
converges. Since $\ln x\ge1$ for $x\ge e$, we have
$$
\frac{\ln x}{x}\ge\frac{1}{x}
$$
However
$$
\int_e^\infty\frac{1}{x}\,dx
$$
does not converge, because $\lim_{x\to \infty}\ln x=\infty$, therefore also the given integral is divergent.
A: Hint. Observe that, for $x>0$,
$$
\frac{d}{dx}\left(\ln x \right)^2=2\frac{\ln x}{x}
$$ and that, as $x \to \infty$,
$$
\left(\ln x \right)^2 \to \infty.
$$
A: An antiderivative of the integrand is:
\begin{align}
\int 81 \frac{\ln(x)}{x} dx 
&= 81 \frac{\ln(x)^2}{2} + C
\end{align}
We have $\ln(1) = 0$ and $\lim_{x\to\infty} \ln(x) = \infty$.
This gives
\begin{align}
\int\limits_1^\infty 81 \frac{\ln(x)}{x} dx &= 
\frac{81}{2}\left[\ln(x)^2\right]_{x=1}^{x \to \infty} \\ 
&= \lim_{x\to\infty} \frac{81}{2} \ln(x)^2 = \infty
\end{align}
A: You can prove that the integral is divergent by Bertrand criterion where here $\alpha=1$ and $\beta=-1.$
